Problem: A $20$ -meter ladder is sliding down a vertical wall so the distance between the top of the ladder and the ground is decreasing at $8$ meters per minute. At a certain instant, the top of the ladder is $12$ meters from the ground. What is the rate of change of the distance between the bottom of the ladder and the wall at that instant (in meters per minute)? Choose 1 answer: Choose 1 answer: (Choice A) A $6$ (Choice B) B $24$ (Choice C) C $\dfrac{32}{3}$ (Choice D) D $8$
Setting up the math Let... $a(t)$ denote the distance between the top of the ladder and the ground at time $t$, $b(t)$ denote the distance between the bottom of the ladder and the wall at time $t$, and $c$ denote the length of the ladder (which is always $20$ meters). $a(t)$ $b(t)$ $c$ We are given that $c=20$ and $a'(t)=-8$ (notice that $a'$ is negative). We are also given that $a(t_0)=12$ for a specific time $t_0$. We want to find $b'(t_0)$. Relating the measures The measures relate to each other through the Pythagorean theorem: $\begin{aligned} [a(t)]^2+[b(t)]^2&=c^2 \\\\\\ [a(t)]^2+[b(t)]^2&=20^2 \end{aligned}$ We can differentiate both sides to find an expression for $b'(t)$ : $b'(t)=-\dfrac{a(t)a'(t)}{b(t)}$ Using the information to solve In order to find $b'(t_0)$ we need to find $b(t_0)$. Using the Pythagorean theorem and the fact that $a(t_0)=12$ and $c=20$, we can find that $b(t_0)=16$. Let's plug ${a(t_0)}={12}$, ${a'(t_0)}={-8}$, and ${b(t_0)}={16}$ into the expression for $b'(t_0)$ : $\begin{aligned} b'(t_0)&=-\dfrac{{a(t_0)}{a'(t_0)}}{{b(t_0)}} \\\\ &=-\dfrac{({12})({-8})}{({16})} \\\\ &=6 \end{aligned}$ In conclusion, the rate of change of the distance between the bottom of the ladder and the wall at that instant is $6$ meters per minute. Since the rate of change is positive, we know that the distance is increasing.